GUJCET 2017 Physics Question Paper with Answer and Solution

(D) The formula for capacitance is $C = \frac{Q}{V}$.
Since potential $V = \frac{W}{Q}$,where $W$ is work and $Q$ is charge,we substitute this into the capacitance formula:
$C = \frac{Q^2}{W}$.
The dimensional formula for work $W$ is $[M^1 L^2 T^{-2}]$.
Substituting the dimensions:
$C = \frac{Q^2}{[M^1 L^2 T^{-2}]}$.
$C = M^{-1} L^{-2} T^2 Q^2$.

(B) The given $AC$ voltage is $V = V_m \cos(\omega t)$, where $V_m = 5 \text{ V}$ and $\omega = 1000 \text{ rad/s}$.
For an $L-R$ circuit, the impedance $Z$ is given by $Z = \sqrt{R^2 + X_L^2}$, where $X_L = \omega L$.
Given $L = 3 \text{ mH} = 3 \times 10^{-3} \text{ H}$ and $R = 4 \text{ } \Omega$.
Calculating inductive reactance: $X_L = 1000 \times 3 \times 10^{-3} = 3 \text{ } \Omega$.
Calculating impedance: $Z = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ } \Omega$.
The maximum current $I_m$ is given by $I_m = \frac{V_m}{Z}$.
Substituting the values: $I_m = \frac{5}{5} = 1 \text{ A}$.

(D) The given equation for the alternating current is $I = I_m \cos(\omega t + \phi)$,where $I_m$ is the peak current.
Comparing this with the given equation $I = 50 \cos(100t + 45^{\circ}) \ A$,we find the peak current $I_m = 50 \ A$.
The root mean square current $I_{rms}$ is related to the peak current $I_m$ by the formula $I_{rms} = \frac{I_m}{\sqrt{2}}$.
Substituting the value of $I_m$,we get $I_{rms} = \frac{50}{\sqrt{2}}$.
To simplify,multiply the numerator and denominator by $\sqrt{2}$: $I_{rms} = \frac{50 \sqrt{2}}{2} = 25 \sqrt{2} \ A$.
Therefore,the correct option is $D$.

(D) The power in an $AC$ circuit is given by the formula: $P = VI \cos \phi$,where $\cos \phi$ is the power factor.
Given values are: $P = 63 \ W$,$V = 210 \ V$,and $I = 3 \ A$.
Rearranging the formula to solve for the power factor: $\cos \phi = \frac{P}{VI}$.
Substituting the values: $\cos \phi = \frac{63}{3 \times 210}$.
$\cos \phi = \frac{63}{630} = 0.1$.
Therefore,the power factor is $0.1$.

(A) For an electron in a hydrogen atom,the relationship between total energy $(E)$,kinetic energy $(K)$,and potential energy $(U)$ is given by:
$E = -K = \frac{U}{2}$
Therefore,the potential energy is $U = 2 \times E$.
Given that the total energy $E = -3.4 \ eV$,we have:
$U = 2 \times (-3.4 \ eV) = -6.8 \ eV$.
Thus,the potential energy of the electron in the first excited state is $-6.8 \ eV$.

(C) The $10 \Omega$ and $40 \Omega$ resistors are connected in parallel. Let $i_1$ be the current through the $10 \Omega$ resistor and $i_2$ be the current through the $40 \Omega$ resistor.
Since they are in parallel,the potential difference across them is the same:
$i_1 \times 10 = i_2 \times 40$
Given $i_1 = 2.5 \text{ A}$,we have:
$2.5 \times 10 = i_2 \times 40$
$i_2 = \frac{25}{40} = 0.625 \text{ A}$
At the junction point,the total current $i$ flowing through the circuit is:
$i = i_1 + i_2 = 2.5 + 0.625 = 3.125 \text{ A}$
The equivalent resistance $R'$ of the parallel combination of $10 \Omega$ and $40 \Omega$ is:
$R' = \frac{10 \times 40}{10 + 40} = \frac{400}{50} = 8 \Omega$
The total resistance of the circuit is $R'' = (R + 8) \Omega$.
Using Ohm's law,$V = i \times R''$:
$50 = 3.125 \times (R + 8)$
$R + 8 = \frac{50}{3.125} = 16$
$R = 16 - 8 = 8 \Omega$
Thus,the value of $R$ is $8 \Omega$.

(D) According to the carbon resistor colour code:
$1$. The first band is Brown,which corresponds to the digit $1$.
$2$. The second band is Red,which corresponds to the digit $2$.
$3$. The third band is Orange,which acts as the multiplier $10^3$.
$4$. The fourth band is Silver,which represents the tolerance of $\pm 10 \%$.
Combining these,the resistance $R$ is given by:
$R = (12 \times 10^3 \pm 10 \%) \ \Omega$
$R = (12 \ k\Omega \pm 10 \%)$
Therefore,the correct option is $(D)$.

(C) According to Gauss's Law,the electric flux $\phi$ linked with a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_0}$,where $q_{enclosed}$ is the net charge enclosed by the surface.
For surface $P$:
The charges enclosed are $+q, -q, -q$.
Net charge $q_{P} = (+q) + (-q) + (-q) = -q$.
Therefore,flux $\phi_{P} = \frac{-q}{\varepsilon_0}$.
For surface $Q$:
The charges enclosed are $+q, -q, +q$.
Net charge $q_{Q} = (+q) + (-q) + (+q) = +q$.
Therefore,flux $\phi_{Q} = \frac{q}{\varepsilon_0}$.
Thus,the flux linked with $P$ and $Q$ are $\frac{-q}{\varepsilon_0}$ and $\frac{q}{\varepsilon_0}$ respectively.

(B) The electric field intensity due to a single thin infinite sheet with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2\varepsilon_0}$.
For two parallel sheets with the same surface charge density $\sigma$:
$1$. Between the sheets: The electric fields due to the two sheets are equal in magnitude but opposite in direction. Thus,the net electric field is $E_{net} = E_1 - E_2 = \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = 0$.
$2$. Outside the sheets: The electric fields due to both sheets point in the same direction. Thus,the net electric field is $E_{net} = E_1 + E_2 = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}$.
Therefore,the electric field intensity between the sheets is $0$ and outside the sheets is $\frac{\sigma}{\varepsilon_0}$.
The correct option is $B$.

(B) The dimension of inductance $L$ is given by $V = L \frac{di}{dt}$,so $[L] = [V][T][I]^{-1}$.
The dimension of resistance $R$ is given by $V = IR$,so $[R] = [V][I]^{-1}$.
Therefore,the dimension of the ratio $\frac{L}{R}$ is:
$\left[ \frac{L}{R} \right] = \frac{[V][T][I]^{-1}}{[V][I]^{-1}} = [T]$.
Thus,$\frac{L}{R}$ has the dimension of time.
Correct option is $B$.

(B) The induced electromotive force $(\varepsilon)$ in a rotating coil is given by the formula: $\varepsilon = N B A \omega \sin(\omega t)$.
For the maximum induced emf $(\varepsilon_{\max})$, we set $\sin(\omega t) = 1$.
Thus, the formula becomes: $\varepsilon_{\max} = N B A \omega$.
Given values:
Number of turns $(N)$ = $100$
Magnetic field strength $(B)$ = $0.3 \ T$
Area $(A)$ = $2.5 \ m^2$
Angular velocity $(\omega)$ = $60 \ rad \ s^{-1}$
Substituting these values into the formula:
$\varepsilon_{\max} = 100 \times 0.3 \times 2.5 \times 60$
$\varepsilon_{\max} = 30 \times 2.5 \times 60$
$\varepsilon_{\max} = 75 \times 60 = 4500 \ V$.
Converting to $kV$:
$\varepsilon_{\max} = 4.5 \ kV$.

(A) The number of waves in a given length $L$ is given by the formula: $\text{Number of waves} = \frac{L}{\lambda}$.
Since the radiation is passing through air,its velocity $v$ is approximately equal to the speed of light $c = 3 \times 10^8 \text{ m/s}$.
Given frequency $\nu = 9 \text{ GHz} = 9 \times 10^9 \text{ Hz}$ and length $L = 1 \text{ m}$.
Using the relation $\lambda = \frac{c}{\nu}$,we substitute this into the formula:
$\text{Number of waves} = \frac{L \times \nu}{c}$.
Substituting the values:
$\text{Number of waves} = \frac{1 \times 9 \times 10^9}{3 \times 10^8} = 3 \times 10^1 = 30$.
Thus,the number of waves is $30$.

(A) The correct answer is $A$. Gamma rays have very high energy and high penetrating power. Due to this property,they are used in medical treatments,specifically in radiotherapy,to kill or destroy cancer cells by damaging their $DNA$.

(A) The work done $W$ in rotating an electric dipole in a uniform electric field from an initial angle $\theta_1$ to a final angle $\theta_2$ is given by the formula:
$W = p E (\cos \theta_1 - \cos \theta_2)$
Given that the dipole is initially parallel to the electric field,the initial angle $\theta_1 = 0^{\circ}$.
The dipole is rotated through $180^{\circ}$,so the final angle $\theta_2 = 180^{\circ}$.
Substituting these values into the formula:
$W = p E (\cos 0^{\circ} - \cos 180^{\circ})$
Since $\cos 0^{\circ} = 1$ and $\cos 180^{\circ} = -1$:
$W = p E (1 - (-1))$
$W = p E (1 + 1)$
$W = 2 p E$
Therefore,the work done is $2 p E$.

(B) The electric field $\vec{E}$ is uniform and directed along the $+X$-direction.
Points $P(0,0)$ and $R(0,2)$ lie on the same plane perpendicular to the electric field lines (the $YZ$-plane). Therefore,$P$ and $R$ lie on the same equipotential surface,which implies $V_P = V_R$.
Electric potential decreases in the direction of the electric field. Since point $Q(2,0)$ is further along the $+X$-direction than point $P(0,0)$,the potential at $P$ must be greater than the potential at $Q$,i.e.,$V_P > V_Q$.
Combining these,we get $V_P = V_R$ and $V_P > V_Q$.

(D) The relative permeability $\mu_r$ of a magnetic material is defined as the ratio of the permeability of the material $\mu$ to the permeability of free space $\mu_0$,given by $\mu_r = \frac{\mu}{\mu_0}$.
Also,the relationship between relative permeability and magnetic susceptibility $\chi_m$ is given by $\mu_r = 1 + \chi_m$.
Equating these two expressions,we get $\frac{\mu}{\mu_0} = 1 + \chi_m$.
Therefore,the permeability of the material is $\mu = \mu_0(1 + \chi_m)$.

(D) The expression $\frac{B^2}{2\mu_0}$ represents the magnetic energy density $(u_B)$ stored in a magnetic field.
Energy density is defined as energy per unit volume.
$u_B = \frac{B^2}{2\mu_0} = \frac{\text{Energy}}{\text{Volume}}$
The dimensional formula for energy is $[M^1 L^2 T^{-2}]$ and for volume is $[L^3]$.
Therefore,the dimensional formula for magnetic energy density is:
$\frac{[M^1 L^2 T^{-2}]}{[L^3]} = [M^1 L^{-1} T^{-2}]$
Thus,the correct option is $D$.

(B) The velocity $v$ of the particle can be resolved into two components: one parallel to the magnetic field $(v_{\parallel} = v \cos \theta)$ and one perpendicular to the magnetic field $(v_{\perp} = v \sin \theta)$.
Due to the parallel component $v_{\parallel}$,the particle moves linearly along the direction of the magnetic field.
Due to the perpendicular component $v_{\perp}$,the magnetic force $F = q(v \times B)$ acts as a centripetal force,causing the particle to move in a circular path in the plane perpendicular to the field.
The combination of linear motion along the field and circular motion perpendicular to the field results in a helical path.

(A) The force $F$ on a charge $q$ moving with velocity $v$ in a magnetic field $B$ is given by $F = qvB \sin \theta$. Since the proton moves perpendicular to the field,$\theta = 90^{\circ}$,so $F = qvB$.
Given kinetic energy $K = 2 \ MeV = 2 \times 10^6 \times 1.6 \times 10^{-19} \ J = 3.2 \times 10^{-13} \ J$.
Using $K = \frac{1}{2}mv^2$,the velocity $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 \times 3.2 \times 10^{-13}}{1.6 \times 10^{-27}}} = \sqrt{4 \times 10^{14}} = 2 \times 10^7 \ m/s$.
Now,$F = (1.6 \times 10^{-19} \ C) \times (2 \times 10^7 \ m/s) \times (2.5 \ T)$.
$F = 1.6 \times 10^{-19} \times 5 \times 10^7 = 8 \times 10^{-12} \ N$.

(C) The magnetic field due to a long straight wire carrying current $I$ at a distance $d$ is given by $B = \frac{\mu_0 I}{2 \pi d}$.
For point $P$,the distance from wire $1$ is $r$. Thus,the magnetic field due to wire $1$ is $B_1 = \frac{\mu_0 I}{2 \pi r}$ (directed into the page).
The distance from wire $2$ is $2r + r = 3r$. Thus,the magnetic field due to wire $2$ is $B_2 = \frac{\mu_0 I}{2 \pi (3r)} = \frac{\mu_0 I}{6 \pi r}$ (directed into the page).
Since both fields are in the same direction (into the page),the net magnetic field $B_{net}$ is:
$B_{net} = B_1 + B_2$
$B_{net} = \frac{\mu_0 I}{2 \pi r} + \frac{\mu_0 I}{6 \pi r}$
$B_{net} = \frac{\mu_0 I}{2 \pi r} (1 + \frac{1}{3}) = \frac{\mu_0 I}{2 \pi r} (\frac{4}{3}) = \frac{2 \mu_0 I}{3 \pi r}$.

(C) The magnifying power $m$ of a simple microscope when the image is formed at infinity (normal adjustment) is given by the formula: $m = \frac{D}{f}$.
Here,$D$ is the near point distance of the normal eye,which is $25 \ cm$.
The focal length of the convex lens is $f = 12.5 \ cm$.
Substituting these values into the formula:
$m = \frac{25}{12.5} = 2$.
Therefore,the magnification is $2$.

(B) In an astronomical telescope,the objective lens forms an image of a distant object at its focal plane.
For normal adjustment,the final image is formed at infinity,and the distance between the objective lens and the eyepiece is $L = f_{0} + f_{e}$.
When the final image is formed at the near point (least distance of distinct vision),the eyepiece is moved closer to the objective lens,but the total length $L$ must be at least $f_{0} + f_{e}$ to accommodate the focal lengths of both lenses in the optical path.
Therefore,the tube length $L$ is generally given by $L \geq f_{0} + f_{e}$.

(A) For an ideal diode,the forward bias resistance is $0 \ \Omega$ and the reverse bias resistance is $\infty \ \Omega$.
$(i)$ Case $V_A > V_B$:
In this case,both diodes $D_1$ and $D_2$ are forward biased.
Therefore,the resistance of each branch is $50 \ \Omega + 0 \ \Omega = 50 \ \Omega$.
Since the two branches are in parallel,the equivalent resistance $R_{AB}$ is given by:
$\frac{1}{R_{AB}} = \frac{1}{50} + \frac{1}{50} = \frac{2}{50} = \frac{1}{25}$
$R_{AB} = 25 \ \Omega$.
(ii) Case $V_B > V_A$:
In this case,both diodes $D_1$ and $D_2$ are reverse biased.
Therefore,the resistance of each branch is $50 \ \Omega + \infty \ \Omega = \infty \ \Omega$.
Since the two branches are in parallel,the equivalent resistance $R_{AB}$ is given by:
$\frac{1}{R_{AB}} = \frac{1}{\infty} + \frac{1}{\infty} = 0 + 0 = 0$
$R_{AB} = \infty \ \Omega$.
Thus,the equivalent resistances are $25 \ \Omega$ and $\infty \ \Omega$ respectively.

(B) For bright fringes in Young's double slit experiment,the condition is given by $d \sin \theta = n \lambda$,where $d$ is the slit separation,$\theta$ is the angle,$n$ is the order of the fringe,and $\lambda$ is the wavelength.
Given that $d = \lambda$,the equation becomes $\lambda \sin \theta = n \lambda$,which simplifies to $\sin \theta = n$.
Since the maximum value of $\sin \theta$ is $1$,we have $n \leq 1$.
The possible integer values for $n$ are $-1, 0, 1$.
Thus,there are $3$ bright fringes in total.